3.273 \(\int \frac{1}{x \sqrt{c+d x^3} (4 c+d x^3)} \, dx\)

Optimal. Leaf size=65 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{3} \sqrt{c}}\right )}{6 \sqrt{3} c^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{6 c^{3/2}} \]

[Out]

-ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])]/(6*Sqrt[3]*c^(3/2)) - ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]]/(6*c^(3/2))

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Rubi [A]  time = 0.0553441, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {446, 86, 63, 208, 203} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{3} \sqrt{c}}\right )}{6 \sqrt{3} c^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{6 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[c + d*x^3]*(4*c + d*x^3)),x]

[Out]

-ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])]/(6*Sqrt[3]*c^(3/2)) - ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]]/(6*c^(3/2))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 86

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), In
t[(e + f*x)^p/(a + b*x), x], x] - Dist[d/(b*c - a*d), Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] &&  !IntegerQ[p]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{x \sqrt{c+d x^3} \left (4 c+d x^3\right )} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x} (4 c+d x)} \, dx,x,x^3\right )\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^3\right )}{12 c}-\frac{d \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x} (4 c+d x)} \, dx,x,x^3\right )}{12 c}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{3 c+x^2} \, dx,x,\sqrt{c+d x^3}\right )}{6 c}+\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^3}\right )}{6 c d}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{3} \sqrt{c}}\right )}{6 \sqrt{3} c^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{6 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0184914, size = 59, normalized size = 0.91 \[ -\frac{\sqrt{3} \tan ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{3} \sqrt{c}}\right )+3 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{18 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[c + d*x^3]*(4*c + d*x^3)),x]

[Out]

-(Sqrt[3]*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])] + 3*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(18*c^(3/2))

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Maple [C]  time = 0.018, size = 433, normalized size = 6.7 \begin{align*}{\frac{{\frac{i}{36}}\sqrt{2}}{{c}^{2}{d}^{2}}\sum _{{\it \_alpha}={\it RootOf} \left ( d{{\it \_Z}}^{3}+4\,c \right ) }{\sqrt [3]{-{d}^{2}c}\sqrt{{{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( -i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}\sqrt{{d \left ( x-{\frac{1}{d}\sqrt [3]{-{d}^{2}c}} \right ) \left ( -3\,\sqrt [3]{-{d}^{2}c}+i\sqrt{3}\sqrt [3]{-{d}^{2}c} \right ) ^{-1}}}\sqrt{{-{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}} \left ( i\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,\sqrt{3}d-i\sqrt{3} \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}+2\,{{\it \_alpha}}^{2}{d}^{2}-\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,d- \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}} \right ){\it EllipticPi} \left ({\frac{\sqrt{3}}{3}\sqrt{{i\sqrt{3}d \left ( x+{\frac{1}{2\,d}\sqrt [3]{-{d}^{2}c}}-{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}},{\frac{1}{6\,cd} \left ( 2\,i\sqrt [3]{-{d}^{2}c}\sqrt{3}{{\it \_alpha}}^{2}d-i \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}\sqrt{3}{\it \_alpha}+i\sqrt{3}cd-3\, \left ( -{d}^{2}c \right ) ^{2/3}{\it \_alpha}-3\,cd \right ) },\sqrt{{\frac{i\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c} \left ( -{\frac{3}{2\,d}\sqrt [3]{-{d}^{2}c}}+{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ) ^{-1}}} \right ){\frac{1}{\sqrt{d{x}^{3}+c}}}}}-{\frac{1}{6}{\it Artanh} \left ({\sqrt{d{x}^{3}+c}{\frac{1}{\sqrt{c}}}} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(d*x^3+4*c)/(d*x^3+c)^(1/2),x)

[Out]

1/36*I/d^2/c^2*2^(1/2)*sum((-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*
c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1
/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3
^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2
)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),1/6/d*(2*I*(-d^2*
c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/c,(I*
3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d+4
*c))-1/6*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (d x^{3} + 4 \, c\right )} \sqrt{d x^{3} + c} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(d*x^3+4*c)/(d*x^3+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((d*x^3 + 4*c)*sqrt(d*x^3 + c)*x), x)

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Fricas [A]  time = 1.53335, size = 392, normalized size = 6.03 \begin{align*} \left [-\frac{2 \, \sqrt{3} \sqrt{c} \arctan \left (\frac{\sqrt{3} \sqrt{d x^{3} + c}}{3 \, \sqrt{c}}\right ) - 3 \, \sqrt{c} \log \left (\frac{d x^{3} - 2 \, \sqrt{d x^{3} + c} \sqrt{c} + 2 \, c}{x^{3}}\right )}{36 \, c^{2}}, -\frac{\sqrt{3} \sqrt{-c} \log \left (\frac{d x^{3} + 2 \, \sqrt{3} \sqrt{d x^{3} + c} \sqrt{-c} - 2 \, c}{d x^{3} + 4 \, c}\right ) - 6 \, \sqrt{-c} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{c}\right )}{36 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(d*x^3+4*c)/(d*x^3+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/36*(2*sqrt(3)*sqrt(c)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c)) - 3*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 +
c)*sqrt(c) + 2*c)/x^3))/c^2, -1/36*(sqrt(3)*sqrt(-c)*log((d*x^3 + 2*sqrt(3)*sqrt(d*x^3 + c)*sqrt(-c) - 2*c)/(d
*x^3 + 4*c)) - 6*sqrt(-c)*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c))/c^2]

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Sympy [A]  time = 12.594, size = 63, normalized size = 0.97 \begin{align*} \frac{\operatorname{atan}{\left (\frac{\sqrt{c + d x^{3}}}{\sqrt{- c}} \right )}}{6 c \sqrt{- c}} - \frac{\sqrt{3} \operatorname{atan}{\left (\frac{\sqrt{3} \sqrt{c + d x^{3}}}{3 \sqrt{c}} \right )}}{18 c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(d*x**3+4*c)/(d*x**3+c)**(1/2),x)

[Out]

atan(sqrt(c + d*x**3)/sqrt(-c))/(6*c*sqrt(-c)) - sqrt(3)*atan(sqrt(3)*sqrt(c + d*x**3)/(3*sqrt(c)))/(18*c**(3/
2))

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Giac [A]  time = 1.11447, size = 72, normalized size = 1.11 \begin{align*} -\frac{\sqrt{3} \arctan \left (\frac{\sqrt{3} \sqrt{d x^{3} + c}}{3 \, \sqrt{c}}\right )}{18 \, c^{\frac{3}{2}}} + \frac{\arctan \left (\frac{\sqrt{d x^{3} + c}}{\sqrt{-c}}\right )}{6 \, \sqrt{-c} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(d*x^3+4*c)/(d*x^3+c)^(1/2),x, algorithm="giac")

[Out]

-1/18*sqrt(3)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c))/c^(3/2) + 1/6*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt
(-c)*c)